$f(x, y) = \cos(x + y) - \sin(x)$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( \dfrac{\pi}{2} + \dfrac \pi 2 k, \dfrac{\pi}{2}+ \dfrac \pi 2 j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$ (Choice B) B $\left( \dfrac{\pi}{2} + \pi k, \dfrac{\pi}{2}+ \pi j \right)$, where $k = \ldots -1, 0, 1 \ldots$ (Choice C) C $\left( \dfrac{\pi}{2} + 2\pi k, \dfrac{\pi}{2}+ 2\pi j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$ (Choice D) D There are no critical points.
A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} -\sin(x + y) - \cos(x) \\ \\ -\sin(x + y) \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} -\sin(x + y) - \cos(x) = 0 \\ \\ -\sin(x + y) = 0 \end{cases}$ Subtracting the first equation from the second, we get $\cos(x) = 0$. There are many values for $x$ that satisfy this equation. For example, $x = \dfrac{\pi}{2}$ makes $\cos(x) = 0$. Note that $\cos(x)$ is generally periodic every $2\pi$, but it crosses the $x$ -axis every $\pi$. So we can add any integer multiple of $\pi$ to get a new solution. All solutions will have the form $x = \dfrac{\pi}{2} + \pi k$, for some integer $k$. Now let's solve $-\sin(x + y) = 0$. Substituting $x$ in and rewriting: $\sin\left( \dfrac{\pi}{2} + \pi k + y \right) = 0$ There are many values for $y$ that satisfy this equation. For example, $y = \dfrac{\pi}{2}$ makes $\sin(x + y) =\sin(\pi+\pi k)= 0$. As before, we can add any integer multiple of $\pi$ to get a new solution. All solutions will have the form $y =\dfrac\pi 2+\pi j$, for some integer $j$. Therefore, $f$ has critical points at $\left( \dfrac{\pi}{2} + \pi k, \dfrac{\pi}{2}+ \pi j \right)$, where $k, j = \ldots -1, 0, 1 \ldots$